Generating all combinations
pataphor
pataphor at gmail.com
Tue Jun 2 08:53:34 CEST 2009
Mensanator wrote:
> I couldn't do that if they weren't subsets.
Right. Sometimes one just has to assume things are different even if
they look the same on the surface. That is because else one wouldn't be
able to produce the other generators. I guess it would also work the
other way around, assuming things are the same even when they look
different.
For example see my two functions:
def repeat_each(seq,n):
while True:
for x in seq:
for i in range(n):
yield x
def repeat_all(seq,n):
while True:
for i in range(n):
for x in seq:
yield x
(I should probably not smoke stuff before posting, but anyway)
They are the same, except for switching two lines of code. But for the
second one ('repeat_all') the argument 'n' seems redundant. What does it
even mean to repeat a sequence n times, and do that forever? Isn't that
the same as just repeating the sequence itself, forever? So that's how
we arrive at itertools.cycle . The second argument is there, but it
would be foolish to include it, so it is left out.
But now let's look at how itertools.repeat should really be. It should
look like 'repeat_each' above. Here second argument ('n') *is*
necessary, or else the generator would just infinitely repeat only the
first element of the sequence, which is obviously nonsense. But that is
exactly why itertools.repeat does not accept a sequence (like cycle, its
virtual brother) but instead it has degenerated into something that just
repeats only one thing n times, which is stupid.
So to set things right one has to forget everything and just write
complete balderdash if necessary, if it only leads to finally
understanding how the two are related. Then one can see that
itertools.repeat should be an infinite generator *on sequences* that
however still needs a second argument specifying how many times each
individual item should be repeated, and that itertools.cycle's second
argument is there but hidden.
P.
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